1. How many grams of P4 will react with 35.0 grams of H2O according to the balanced equation P4+5O2+6H2O=4H3PO4?
2. If you have 50 grams of P4, 40 grams of H2O, and an unlimited supply of O2, which reactant (according to the above balanced equation) is the limiting reactant? How much H3PO4 can be produced based upon the amounts of reactants?
Limiting reagent problems are those in which the amount of more than one reactant is given. I solve these problems by solving two stoichiometry problems. Here is an example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Solve once using reactant 1 assuming you have all of reactant 2 needed; then solve again using reactant 2 and assuming you have all of reactant 1 needed. You will obtain two answers; of course both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
Post your work if you get stuck.
To answer both questions, we need to use the concept of stoichiometry and the balanced equation provided.
1. To determine the grams of P4 that will react with 35.0 grams of H2O, we first need to identify the molar mass of each substance.
- Molar mass of P4 (Phosphorus): 4 * atomic mass of P (Phosphorus)
- Molar mass of H2O (Water): 2 * atomic mass of H (Hydrogen) + atomic mass of O (Oxygen)
Next, we convert the grams of H2O to moles using the molar mass of water. The molar mass of water is calculated as follows:
H2O: (2 * 1.008 g/mol) + 16.00 g/mol = 18.016 g/mol
To convert grams to moles, we use the formula:
moles = mass / molar mass
Therefore, moles of H2O = 35.0 g / 18.016 g/mol ≈ 1.943 moles
Based on the balanced equation, we can see that the molar ratio between P4 and H2O is 1:6. This means that for every 6 moles of H2O, 1 mole of P4 is required.
Using this ratio, we can calculate the moles of P4 needed:
moles of P4 = moles of H2O / 6 = 1.943 moles / 6 ≈ 0.324 moles
Finally, we convert moles of P4 to grams using the molar mass of phosphorus (P):
mass = moles * molar mass
mass of P4 = 0.324 moles * (4 * atomic mass of P) = 0.324 moles * (4 * 30.974 g/mol) ≈ 40.19 grams
Therefore, approximately 40.19 grams of P4 will react with 35.0 grams of H2O.
2. To determine the limiting reactant and the amount of H3PO4 formed, we need to calculate the moles of each reactant.
First, we convert the grams of P4 and H2O to moles using their respective molar masses.
moles of P4 = 50 g / (4 * atomic mass of P)
moles of H2O = 40 g / (2 * atomic mass of H + atomic mass of O)
Next, we use the balanced equation to determine the molar ratio between P4 and H2O to identify the limiting reactant.
From the balanced equation: P4 + 5O2 + 6H2O → 4H3PO4
The molar ratio between P4 and H2O is 1:6.
We compare the moles of P4 and H2O to determine the limiting reactant. The reactant with the smaller moles-to-ratio value will be the limiting reactant.
If the calculated moles of P4 divided by 1 (the ratio in the balanced equation) is smaller than the moles of H2O divided by 6 (the ratio in the balanced equation), then P4 is the limiting reactant. Otherwise, H2O is the limiting reactant.
Finally, to determine the amount of H3PO4 produced, we need to calculate the moles of H3PO4 formed based on the limiting reactant.
Using the mole ratio from the balanced equation: 1:4 (1 mole of P4 produces 4 moles of H3PO4)
moles of H3PO4 formed = moles of limiting reactant * (4 moles of H3PO4 / 1 mole of P4)
Finally, to get the mass of H3PO4 formed, we multiply the moles of H3PO4 by its molar mass.
mass of H3PO4 formed = moles of H3PO4 formed * molar mass of H3PO4
Remember to use the molar masses of P4 (Phosphorus) and H3PO4 (Phosphoric Acid) for the calculations.
Note: The limiting reactant is the reactant that, after being used up completely, limits the amount of product that can be formed.